Introduce a battery to an LR circuit

[laser]

Homework Statement

Introduce a battery to an LR circuit

Relevant Equations

none

Let's say I have an LR circuit. I understand the mathematical derivation of this but would like to understand it conceptually.

Okay, so a battery is introduced (let's say 12 V). The inductor hates the change, so it induces 12 V in the opposite direction, opposing the battery.

So what happens then? IDK! My guess is that the two emfs cancel out, (which seems reasonable), and that means current is 0. Okay, fine, the book agrees with me. But what happens next? I know what is supposed to happen - the induced emf goes from 12 -> 0 in an exponential fashion. But I don't know *why*.

To me, it would seem reasonable for the induced emf to go to zero. Because di/dt = 0. But instead it decreases exponentially. So I must have a flaw in my understanding somewhere. Thanks!

 

[BvU]

Look at the current! It wants to jump to V/R and can't do so instantaneously because of the emf

$\$

 

[laser]

Look at the current! It wants to jump to V/R and can't do so instantaneously because of the emf

$\$

Ya but my question was that why doesn't the induced emf go to zero. Because di/dt is zero.

 

[berkeman]

The inductor hates the change

Never anthropomorphize inductors. They hate it when you do that.

Let's say I have an LR circuit. I understand the mathematical derivation of this but would like to understand it conceptually.

For me, the conceptual understanding comes from the differential equation relating an inductor's voltage to its current. I'm not sure there is another way to try to understand that behavior (unless you want to try to convert it into puppy dogs and kitty cats or something). Do you understand this equation?

$$v(t) = L \frac{di(t)}{dt}$$

 

[Herman Trivilino]

Ya but my question was that why doesn't the induced emf go to zero. Because di/dt is zero.

After a long time the voltage across the (ideal) inductor goes to zero. The current through the inductor is steady ($\frac{di}{dt}=0$), but the voltage $V$ across the battery is not zero, the current is $\frac{V}{R}$.

 

[laser]

After a long time the voltage across the (ideal) inductor goes to zero. The current through the inductor is steady ($\frac{di}{dt}=0$), but the voltage $V$ across the battery is not zero, the current is $\frac{V}{R}$.

But if di/dt = 0, then the induced emf = 0 too? As induced emf = L*di/dt. Do you see my point?

 

[laser]

Never anthropomorphize inductors. They hate it when you do that.


For me, the conceptual understanding comes from the differential equation relating an inductor's voltage to its current. I'm not sure there is another way to try to understand that behavior (unless you want to try to convert it into puppy dogs and kitty cats or something). Do you understand this equation?

$$v(t) = L \frac{di(t)}{dt}$$

Yes, the maths behind deriving this makes sense to me. But not the conceptual understanding. Because if di/dt = 0, then surely induced emf = 0 too! But no, the induced emf exponentially shrinks to 0...

 

[Herman Trivilino]

But if di/dt = 0, then the induced emf = 0 too? As induced emf = L*di/dt. Do you see my point?

Yes. The induced emf, in other words the voltage across the (ideal) inductor is zero. But there is a nonzero voltage (emf) across the battery.

 

[Herman Trivilino]

Yes, the maths behind deriving this makes sense to me. But not the conceptual understanding. Because if di/dt = 0, then surely induced emf = 0 too! But no, the induced emf exponentially shrinks to 0...

But that happens only after a long time, when the current becomes steady. In other words $\frac{di}{dt}$ exponentially decreases to zero.

 

[laser]

Yes. The induced emf, in other words the voltage across the (ideal) inductor is zero. But there is a nonzero voltage (emf) across the battery.

I might have misphrased my original question...

"So what happens then? IDK! My guess is that the two emfs cancel out, (which seems reasonable), and that means current is 0. Okay, fine, the book agrees with me. But what happens next? I know what is supposed to happen - the induced emf goes from 12 -> 0 in an exponential fashion. But I don't know *why*.

To me, it would seem reasonable for the induced emf to go to zero. Because di/dt = 0. But instead it decreases exponentially. So I must have a flaw in my understanding somewhere. Thanks!"

I meant to say it would seem reasonable for the induced emf to go to zero IMMEDIATELY. But it doesn't. Does this help to clarify my question?

i.e. At the beginning, 12 V is induced, so no current is flowing. So di/dt is 0. Therefore, no emf should be induced. HOWEVER, emf is still induced, e.g. it would be 11 V, and this decreases to 0.

 

[Herman Trivilino]

so no current is flowing. So di/dt is 0.

No. You can have $i$ zero and $\frac{di}{dt}$ nonzero.

 

[Herman Trivilino]

Have you tried re-reading the $RL$ section in your textbook?

 

[DaveE]

Perhaps you are misunderstanding or misusing the concept of induced emf. "Induced voltage" is really only useful if you have an external magnetic field imposed, like in a transformer.

In your example the battery is imposing a voltage across the inductor which will cause current to increase, according to the inductor equation. That current will create an additional voltage drop across the series resistance which will in turn decrease the voltage across the inductor.

In the inductor equation $V=L \frac{di}{dt}$ it doesn't matter what words you use to describe the voltage or where it comes from. This is what inductors do regardless.

https://www.khanacademy.org/science...-and-forced-response/a/ee-rl-natural-response

 

[Steve4Physics]

I meant to say it would seem reasonable for the induced emf to go to zero IMMEDIATELY. But it doesn't. Does this help to clarify my question?

i.e. At the beginning, 12 V is induced, so no current is flowing. So di/dt is 0.

I'll give it a go too! See if this helps.

The battery is connected at $t=0$. In the first tiny time-interval, from $t= 0$ to $t = \delta t$, the current changes from 0 to $\delta I$.

(We should really use the differentials $dt$ and $dI$ which are the limits taking $\delta t \rightarrow 0$ and $\delta I \rightarrow 0$. But it might help to understand the process better by first considering finite (but tiny) differences.)

The inductor initially produces an emf: $|L \frac {\delta I}{\delta t}| = 12V$ (approx.).

As time progress the current rises and asymptotically approaches a steady value and $|L \frac {\delta I}{\delta t}|$ falls and asymptotically approaches zero.

 

[Herman Trivilino]

At the beginning, 12 V is induced, so no current is flowing. [...]. Therefore, no emf should be induced. HOWEVER, emf is still induced

The emf across the battery is 12 volts. As @Steve4Physics explains in Post #14 the emf across the inductor is 12 volts. The two emf's are in opposite directions so the total is zero, thus the current $i$ is zero. But $\frac{di}{dt}$ is NOT zero, in fact it has its maximum value!

 

[laser]

Thanks guys, for some silly reason I got mixed up with di/dt and i! The general pattern of the decrease in the induced emf makes sense now, and the fact that the decrease is exponential is from the DE.

 

[Herman Trivilino]

Thanks guys, for some silly reason I got mixed up with di/dt and i!

That's an easy mistake to make, and a lesson you won't soon, if ever, forget.

The general pattern of the decrease in the induced emf makes sense now, and the fact that the decrease is exponential is from the DE.

Actually, it's from Nature. The DE just describes the (idealized) behavior.